Length and eigenvalue equivalence
C. J. Leininger∗, D. B. McReynolds†, W. D. Neumann‡and A. W. Reid§
December 10, 2006
1
Introduction
Let M be a compact Riemannian manifold, and let ∆ = ∆M denote the Laplace–
Beltrami operator of M acting on L2 (M). The eigenvalue spectrum E (M) consists of the eigenvalues of ∆ listed with their multiplicities. Two manifolds M1
and M2 are said to be isospectral if E (M1 ) = E (M2 ). Geometric and topological
constraints are forced on isospectral manifolds; for example if the manifolds are
hyperbolic (complete with all sectional curvature equal to −1) then they must have
the same volume [18], and so for surfaces the same genus.
Another invariant of M is the length spectrum L (M) of M; that is the set of
all lengths of closed geodesics on M counted with multiplicities. Two manifolds
M1 and M2 are said to be iso-length spectral if L (M1 ) = L (M2 ). Under the hypothesis of negative sectional curvature the invariants E (M) and L (M) are closely
related. For example, it is known that E (M) determines the set of lengths of closed
geodesics, and in the case of closed hyperbolic surfaces, the stronger statement that
E (M) determines L (M) and vice-versa holds [7, 8].
In this paper we address the issue of how much information is lost by forgetting multiplicities. More precisely, for a compact Riemannian manifold M, define
the eigenvalue set (resp. length set and primitive length set) to be the set of
eigenvalues of ∆ (resp. set of lengths all closed geodesics and lengths of all primitive closed geodesics) counted without multiplicities. These sets will be denoted
E(M), L(M) and L p (M) respectively. Two manifolds M1 and M2 are said to be
eigenvalue equivalent (resp. length equivalent and primitive length equivalent)
if E(M1 ) = E(M2 ) (resp. L(M1 ) = L(M2 ) and L p (M1 ) = L p (M2 )). Although length
∗ Partially
supported by the N. S. F.
supported by a C.M.I. lift-off
‡ Partially supported by the N. S. F.
§ Partially supported by the N. S. F.
† Partially
1
spectrum and primitive length spectrum determine each other, the corresponding
statement for length sets is false. Primitive length equivalent manifolds are clearly
length equivalent, but we shall see that the converse is false.
We will focus mainly on hyperbolic manifolds of finite volume. Even in this
setting little seems known about the existence of manifolds which are eigenvalue
(resp. length or primitive length) equivalent but not isospectral or iso-length spectral. Examples of non-compact arithmetic hyperbolic 2–manifolds that are length
equivalent were constructed in Theorem 2 of [24] using arithmetic methods. However, as far as the authors are aware, no examples of closed hyperbolic surfaces that
are length equivalent and not iso-length spectral were known, and it would appear
that no examples of eigenvalue equivalent or primitive length equivalent hyperbolic
manifolds which are not isospectral or iso-length spectral were known. Our main
results rectify this situation for hyperbolic surfaces and indeed for all finite volume
hyperbolic m–manifolds.
Theorem 1.1. Let M be a closed hyperbolic m–manifold. Then there exist infinitely
many pairs of finite covers {M j , N j } of M such that
(a) E(M j ) = E(N j ),
(b) vol(M j )/ vol(N j ) → ∞.
Moreover, E(M j ) = E(N j ) for any Riemannian metric on M.
The method of proof of Theorem 1.1 does not provide (primitive) length equivalent pairs of covers. However, we can prove an analogue for primitive length
equivalence (and hence also length equivalence).
Theorem 1.2. Let M be a finite volume hyperbolic m–manifold. Then there exist
infinitely many pairs of finite covers {M j , N j } of M such that
(a) L p (M j ) = L p (N j )
(b) vol(M j )/ vol(N j ) → ∞.
Moreover, L p (M j ) = L p (N j ) for any Riemannian metric on M.
Indeed, as we point out in §5.1, for every finite volume hyperbolic n–manifold
where n 6= 3, 4, 5 we can produce pairs of finite sheeted covers of arbitrarily large
volume ratio that are both primitive length equivalent and eigenvalue equivalent.
The methods of the paper are largely group theoretic, relying on the fundamental group rather than the geometry, and a quick way to provide lots of examples
in many more situations is given by the following. Recall that a group Γ is called
large if it contains a finite index subgroup that surjects a free non-abelian group.
2
Theorem 1.3. (Theorem 3.1 is a stronger version.) Let M be a compact Riemannian manifold with large fundamental group. Then there exist infinitely many pairs
of finite covers {M j , N j } of M such that
(a) L(M j ) = L(N j ),
(b) E(M j ) = E(N j ),
(c) vol(M j )/ vol(N j ) → ∞.
Moreover, (a) and (b) hold for any Riemannian metric on M, and if π1 (M) is hyperbolic, L p (M j ) = L p (N j ) also holds for any Riemannian metric on M.
Our arguments start with Sunada’s construction [28] of isospectral manifolds,
which was based on a well known construction in number theory of “arithmetically equivalent” number fields (see [17]). Our length equivalence of manifolds
similarly has a number theoretic counterpart called “Kronecker equivalence” of
number fields, as we discovered after doing this work; see the book [10]. The results contained here can thus be viewed as providing the geometric investigation
proposed in the sentence from the last paragraph of that book: “In view of the
relations between arithmetical and Kronecker equivalence, one should also study
Kronecker equivalence in this geometric situation.”
In the final section we collate some remarks and questions. In particular, we
note that Mark Kac’s famous paper “Can one hear the shape of a drum” [9] has
been a catalyst for much of the work on isospectrality, and we revisit that paper
and the Gordon-Webb-Wolpert answer to his question [5] in the light of our work.
2
Equivalence
We first recall Sunada’s construction.
For any finite group G and subgroups H and K of G, we say that H and K are
almost conjugate (or “Gassmann equivalent” in the terminology of Perlis [17]) if
for any g in G the following condition holds (where (g) denotes conjugacy class):
|H ∩ (g)| = |K ∩ (g)| .
In [28] Sunada proved the following theorem relating almost conjugate pairs with
isospectral covers.
Theorem (Sunada). Let M be a closed Riemannian manifold, G a finite group,
and H and K almost conjugate subgroups of G. If π1 (M) admits a homomorphism
onto G, then the finite covers MH and MK associated to the pullback subgroups
of H and K are isospectral. Moreover, the manifolds MH and MK are iso-length
spectral.
3
The proof of this is an easy exercise, but checking when manifolds produced
by Sunada’s method are non-isometric requires more work. However, for length
equivalence far less is required, and the resolution of the isometry problem is built
into our construction.
Length equivalence, primitive length equivalence, and eigenvalue equivalence
each require a different condition on the group G. In each instance, we describe
a group theoretic condition, and then explain how it is used to produce examples
with the desired features.
2.1
Length and primitive length equivalence
Though it is not essential, the group G will always be finite in what follows.
Definition 2.1 (Elementwise conjugacy). Subgroups H and K of G are said to be
elementwise conjugate if for any g in G the following condition holds:
H ∩ (g) 6= 0/
if and only if
(Or, more briefly, H G = K G , where H G =
K ∩ (g) 6= 0.
/
(1)
−1
g∈G g Hg.)
S
It is immediate from the definition that almost conjugate subgroups are elementwise conjugate.
To produce primitive length equivalent manifolds, we impose further conditions on H and K, and also on π1 (M).
Definition 2.2 (Primitive). We shall call a subgroup H of G primitive in G if the
following holds:
(a) All non-trivial cyclic subgroups of H have the same order p (necessarily
prime).
(b)
−1
g∈G g Hg
T
= {1}.
Theorem 2.3. Let M be a Riemannian manifold, G a group, and H and K elementwise conjugate subgroups of G.
(1) If π1 (M) admits a homomorphism onto G, then the covers MH and MK associated to the pullback subgroups of H and K are length equivalent.
(2) If, in addition, H and K are primitive in G and π1 (M) has the property that
any pair of distinct maximal cyclic subgroups of Γ intersect trivially, then
the covers MH and MK associated to the pullback subgroups of H and K are
primitive length equivalent.
4
Remark. It is well known that when M admits a metric of negative sectional
curvature, then π1 (M) satisfies the condition needed to apply Theorem 2.3.
Proof of theorem. To prove (1) it suffices to show that a closed geodesic γ on M
has a lift to a closed geodesic on MH if and only if it has a lift to a closed geodesic
on MK . Let ρ denote the homomorphism π1 (M) → G. By standard covering space
theory, γ has a closed lift to MH if and only if ρ([γ]) ∈ G is conjugate into H. By
assumption this is true for H if and only if it is true for K, proving (1).
For (2) we will show the inclusion L p (MH ) ⊆ L p (MK ); the reverse inclusion
then follows by symmetry. We argue by contradiction, assuming there is a primitive
γ in π1 (MH ), every conjugate of which in π1 (MK ) is imprimitive. Let γK be any
conjugate of γ in π1 (MK ) and let δ ∈ π1 (MK ) and r > 1 be such that δ r = γK . The
arguments splits into two cases.
Case 1. ρ(δ ) = 1.
Since ker ρ < π1 (MH ), all conjugates of δ are contained in π1 (MH ). This contradicts the primitivity of γ, as a π1 (M)–conjugate of δ powers to γ.
Case 2. ρ(δ ) 6= 1, so ρ(δ ) has prime order p by Definition 2.2 (a). We split
this into two subcases.
Case 2.1. ρ(γK ) 6= 1.
Since hρ(γK )i is nontrivial and contained in hρ(δ )i which has prime order, it is
equal to hρ(δ )i. Thus, ρ(µδ µ −1 ) ∈ hρ(γ)i, where µ is the element of π1 (M) conjugating γK to γ. Since the cyclic subgroup hρ(γ)i is contained in H, ρ(µδ µ −1 ) is
contained in H, so µδ µ −1 is an element of π1 (MH ). This contradicts the primitivity
of γ since (µδ µ −1 )r = γ.
Case 2.2. ρ(γK ) = 1.
By Definition 2.2 (b), there exists an element g in G which conjugates ρ(δ )
outside of K. For any element σ ∈ ρ −1 (g), σ γK σ −1 ∈ π1 (MK ) and by assumption
this cannot be primitive. Therefore, there exists δ1 in π1 (MK ) and s > 1
such that
σ γK σ −1 = δ1s . We have the equality (σ δ σ −1 )r = δ1s . By assumption, σ δ σ −1
and hδ1 i are contained in a common maximal cyclic subgroup C of π1 (M). The
intersection of ρ(C) with K is a cyclic subgroup which contains the image of
hρ(δ1 )i. By Definition 2.2 (a), the cyclic subgroups of K have prime order p,
and so |ρ(C) ∩ K| = 1 or p.
Assume first that the latter holds.
Now ρ(C)
has a unique cyclic subgroup of
−1
order p, so ρ(C) ∩ K must equal ρ(σ δ σ ) . Hence the element ρ(σ δ σ −1 ) is in
K, which contradicts the choice of σ . Therefore, we can assume that ρ(C) ∩ K = 1.
Then ρ(δ1 ) = 1. Replacing γK by σ γK σ −1 and δ by δ1 , Case 1 provides the desired
contradiction.
5
2.2
Eigenvalue equivalence
To give context to our construction below of eigenvalue equivalence, we begin by
recalling the following well known equivalent formulation of almost conjugacy.
Proposition 2.4. Subgroups H and K of a finite group G are almost conjugate if
and only if for every finite dimensional complex representation ρ of G,
dim Fix(ρ|H) = dim Fix(ρ|K) ,
where Fix(ρ|H) denotes the subspace of ρ(H)–fixed vectors.
Proof. A convenient reference for the character theory used here and later is [26].
H ) of the trivial character
The dimension of Fix(ρ|H) is the inner product (χ1H , χρ|H
1
H (h), and since
on H and the character of ρ|H. By definition this is |H|
∑h∈H χρ|H
1
characters are constant on conjugacy classes, this equals |H|
∑ |(g) ∩ H| χρG (g),
where the sum is over conjugacy classes in G. Thus the equality dim Fix(ρ|H) =
dim Fix(ρ|K) is equivalent to
1
1
|(g) ∩ H|χρG (g) =
|(g) ∩ K|χρG (g) .
|H| ∑
|K| ∑
(2)
Clearly, almost conjugacy of H and K implies this equality.
For the converse, note first that the equality dim Fix(ρ|H) = dim Fix(ρ|K)
applied to the regular representation of G becomes [G : H] = [G : K], whence
|H| = |K|. Since characters of irreducible representations form a basis for class
functions on G, letting ρ run over all irreducible representations of G in equation
(2) now implies that |(g) ∩ H| = |(g) ∩ K| for each conjugacy class (g).
Definition 2.5. We say subgroups H and K of a finite group G are fixed point
equivalent if for any finite dimensional complex representation ρ of G, the restriction ρ|H has a nontrivial fixed vector if and only if ρ|K does.
Theorem 2.6. Let H and K be fixed point equivalent subgroups of a finite group
G. If M is a compact Riemannian manifold and π1 (M) admits a homomorphism
onto G, then the covers MH and MK associated to the pullbacks in π1 (M) of H and
K are eigenvalue equivalent.
e be the cover of M associated to the pullback in π1 (M) of the trivial
Proof. Let M
e is by isometries, and the quotients by H and
subgroup of G. The action of G on M
e
e → MK , respectively.
K give covering maps pH : M → MH and pK : M
e whose
The covering projection induces an embedding p∗H : L2 (MH ) → L2 (M)
2 e H
2 e
∗
∗
image is the H–fixed subspace L (M) of L (M). Since ∆Me ◦ pH = pH ◦ ∆MH , the
6
e restricts to an action on the λ –eigenspace L2 (M)
e λ and p∗H
action of H on L2 (M)
2
2 e
H
identifies L (MH )λ with (L (M)λ ) . Thus λ is an eigenvalue for MH if and only
e λ )H has positive dimension.
if (L2 (M)
e λ decomposes as a diSince G is finite, the representation of G on L2 (M)
rect sum of finite dimensional representations (in fact, compactness of M implies
e λ is finite dimensional, but we do not need this). Hence, if H and K are
L2 (M)
e λ )H will be non-trivial if and only if (L2 (M)
e λ )K is
fixed point equivalent, (L2 (M)
non-trivial.
Remark. 1. The compactness assumption on M is not necessary. If M is noncompact our argument extends easily to show that under the conditions of the theorem both the discrete and non-discrete spectra of MH and MK agree when viewed
as sets.
2. What makes the Sunada construction work for both the length and eigenvalue spectra is the equivalence of almost conjugacy with the condition of Proposition 2.4. Our weakening of almost conjugacy to elementwise conjugacy on the one
hand, and, via Proposition 2.4, to fixed point equivalence on the other, go in dual
directions. They therefore cannot be expected to be equivalent, and it is a little
surprising that in the examples we know, the two weaker conditions still tend to
have significant overlap.
2.3
Examples
An elementary example of elementwise conjugacy is the following. Let G be the
alternating group Alt(4), and set a = (12)(34) and b = (14)(23). Then the subgroups H = {1, a} and the Klein 4–group K = {1, a, b, ab} are elementwise conjugate. However note that K is not primitive since it is a normal subgroup. In
addition H and K are not fixed point equivalent since K has no fixed vector under
the irreducible 3–dimensional representation of G while H has a fixed vector. On
the other hand, it is not hard to check that H is fixed point equivalent to the trivial
subgroup {1}.
We now generalize this example.
Let F p be the prime field of order p, and let n ≥ 2 be a positive integer. The
n–dimensional special F p –affine group is the semidirect product Fnp o SL(n; F p )
defined by the standard action of SL(n; F p ) on Fnp . We call any F p –vector subspace
V of Fnp a translational subgroup of Fnp o SL(n; F p ).
Theorem 2.7. Let V and W be translational subgroups of G = Fnp o SL(n; F p ).
Then,
(i) if V and W are both non-trivial then they are elementwise conjugate in G,
and they are moreover primitive if they are proper subgroups of Fnp ;
7
(ii) if V and W are both proper subgroups of Fnp then they are fixed point equivalent in G.
Proof. (i). Since SL(n; F p ) acts transitively on non-trivial elements of Fnp , the elementwise conjugacy is immediate. Moreover, conditions (a) and (b) of Definition
2.2 clearly hold for V if V is a proper subgroup of Fnp .
(ii). It suffices to show that any proper translational subgroup V is fixed point
equivalent to the trivial subgroup. So we must show that for any m-dimensional
representation ρ of Fnp o SL(n; F p ) with m > 0, the restriction ρ|V has a nontrivial
fixed subspace when restricted to V . To this end, let χ be the character of ρ.
The dimension of the fixed space of ρ|V is dim(Fix(ρ|V )) = |V1 | ∑v∈V χ(v). Since
χ(1) = m and any two nontrivial elements of V are conjugate in Fnp o SL(n; F p ),
we can rewrite this:
dim(Fix(ρ|V )) =
1
(m + (|V | − 1)χ(x)) ,
|V |
where x ∈ V − {0}. Similarly, the dimension of the fixed space for the full translation subgroup Fnp is
1
dim(Fix(ρ|Fnp )) = n m + (Fnp − 1)χ(x) .
Fp
Thus m + (Fnp − 1)χ(x) ≥ 0, so χ(x) ≥
−m
. Hence,
1
1
dim(Fix(ρ|V )) = |V | (m + (|V | − 1)χ(x)) ≥ |V | m − m
3
|Fnp |−1
|V |−1
|Fnp |−1
> 0.
Proofs of main results
The following is a stronger version of Theorem 1.3:
Theorem 3.1. Let M be a compact Riemannian manifold whose fundamental group
is large. For every integer n ≥ 2 and every odd prime p, there exists a finite tower
of covers of M
M0 −→ M1 −→ . . . −→ Mn−1 −→ Mn −→ M ,
with each Mi → Mi+1 of degree p, such that:
(a) L(M j ) = L(Mk ) for 0 ≤ j, k ≤ n − 1;
(b) E(M j ) = E(Mk ) for 1 ≤ j, k ≤ n;
8
Moreover, (a),(b) hold for any Riemannian metric on M. Finally, if π1 (M) is hyperbolic then for any Riemannian metric on M,
(c) L p (M j ) = L p (Mk ) for 1 ≤ j, k ≤ n − 1.
Proof. Since M is large we can find finite index subgroups which surject any
finitely generated free group, so there is a finite cover X of M with
π1 (X)
/ / Fnp o SL(n; F p ) .
Consider any complete F p –flag
{0} = V0 ⊂ V1 ⊂ V2 ⊂ . . . ⊂ Vn−1 ⊂ Vn = Fnp
in Fnp . Pulling these subgroups back to π1 (X) ⊂ π1 (M) we obtain a tower
M0 −→ M1 −→ . . . −→ Mn−1 −→ Mn −→ M
of corresponding covers of M. The theorem then follows from Theorem 2.7 combined with Theorems 2.3 and 2.6.
This theorem implies Theorems 1.1, 1.2, and 1.3 in the case of closed hyperbolic surfaces. In addition, it is well-known that closed and finite volume hyperbolic manifolds whose fundamental groups are large exist in all dimensions (see
e.g., [12]). This provides examples of hyperbolic manifolds in all dimensions satisfying the conclusions of Theorems 1.1, 1.2, and 1.3. To prove that any closed or
finite volume hyperbolic manifold has finite sheeted covers with these properties
requires additional work.
We mention in passing that Theorem 3.1 (b) applied to surfaces produces arbitrarily long towers of abelian covers
M1 −→ . . . −→ Mn−1 −→ Mn
whose first nontrivial eigenvalue remains constant. On the other hand, it is well
known that any infinite tower of abelian covers of a fixed hyperbolic surface has λ1
tending to zero (see [1] and [28]).
3.1
More families
Theorem 3.2.
(i) Let p > 3 be a prime. Then PSL(2; Z/p2 Z) contains subgroups K < H with [H : K] = p which are fixed point equivalent.
9
(ii) Let k 6= Q be a number field with ring of integers Ok . Let P be the set of
non-dyadic prime ideals p of Ok for which Ok /p = Fq is a non-prime field
(this set is infinite by the Cebotarev Density Theorem). Then for p in P the
group PSL(2; Ok /p2 ) contains subgroups K < H with [H : K] = p which are
primitive and elementwise conjugate in PSL(2; Ok /p2 ).
The proof of Theorem 3.2 will be deferred until §4. Assuming this we complete
the proofs of Theorems 1.1 and 1.2 in the next subsection.
3.2
Completion of proofs
We shall need the following special case of the Strong Approximation Theorem
(see [30] and [16]; see also [11] for a discussion of the proof in the particular case
of hyperbolic manifolds). Suppose M m is a finite volume hyperbolic manifold with
m ≥ 3. We shall identify Isom(Hm ) with PO0 (m, 1) so π1 (M) < PO0 (m, 1). We
can assume there is a number field k such that π1 (M) < PO0 (m, 1; S) for a finite
extension ring S of Ok with k the field of fractions of S (see [22] for the details).
We choose k minimal.
Theorem 3.3 (Strong Approximation). For all but finitely many primes p of S the
image of π1 (M) under the reduction homomorphism
rp j : PO0 (m, 1; S) −→ PO(m, 1; S/p j )
contains the commutator subgroup Ω(m, 1; S/p j ) of PO(m, 1; S/p j ) for all j ≥ 1.
Proof of Theorem 1.1. Theorem 1.1 is shown for hyperbolic surfaces in the comment following the proof of Theorem 3.1, so we can assume that M is a closed
hyperbolic manifold of dimension m ≥ 3. We will produce surjections of π1 (M)
onto finite groups containing PSL(2; Z/p2 Z) for infinitely many p.
Let S be as in the Strong Approximation Theorem above. For all but a finite
number of primes p of S the image of π1 (M) < PO0 (m, 1; S) under the restriction
homomorphism rp j contains Ω(m, 1; S/p j ) for all j ≥ 1. If p is the integer prime
that p divides then rp j (π1 (M)) therefore contains the subgroup Ω(m, 1; Z/p j Z),
and therefore also the subgroup Ω(2, 1; Z/p j Z) of Ω(m, 1; Z/p j Z).
We claim the finite groups Ω(2, 1; Z/p j Z) and PSL(2; Z/p j Z) are isomorphic.
To see this, first recall that the p–adic Lie groups PSL(2; Qp ) and Ω(2, 1; Qp ) are
isomorphic (as p–adic Lie groups). The groups PSL(2; Zp ) and Ω(2, 1; Zp ) are
the respective maximal compact subgroups of PSL(2; Qp ) and Ω(2, 1; Qp ), and are
unique up to isomorphism (see [19, Ch 3.4]). Hence the groups PSL(2; Zp ) and
10
Ω(2, 1; Zp ) are isomorphic as p–adic Lie groups. Reducing modulo the ideal generated by the jth power of the uniformizer π of Zp yields the asserted isomorphism
between Ω(2, 1; Z/p j Z) and PSL(2; Z/p j Z).
Restricting now to j = 2 we have shown rp2 (π1 (M)) contains a subgroup isomorphic to PSL(2; Z/p2 Z). So by passage to a subgroup of finite index in π1 (M),
we can arrange a finite cover X of M with a surjection π1 (X) → PSL(2; Z/p2 Z).
The existence of pairs {M j , N j } as stated in Theorem 1.1 now follows from Theorem 2.6, Theorem 3.2(i), and the infinitude of P.
Proof of Theorem 1.2. We have already shown this for hyperbolic surfaces in the
comments following Theorem 3.1. We next consider hyperbolic 3–manifolds. Let
M be a hyperbolic 3–manifold with holonomy representation π1 (M) < PSL(2; S)
(again S chosen minimally). The field of fractions of S, a finite ring extension of
Z, is necessarily a proper extension of Q, see [14]. In particular, by the Cebotarev
Density Theorem, there exist infinitely many prime ideals p of S such that S/p is
a nontrivial extension of F p . The Strong Approximation Theorem applies here to
see that for all but finitely many among this infinite set of prime ideals of S the
reduction maps
rp2 : π1 (M) −→ PSL(2; S/p2 ),
are onto. By Theorem 3.2 (ii) and Theorem 2.3 (ii), there exists a pair of covers
Np −→ Mp −→ M
with L p (Np ) = L p (Mp ) and vol(Np )/ vol(Mp ) = p.
We extend this to all hyperbolic m–manifolds with m > 3 as follows. Let S =
Z[i] and let P be the set of prime ideals defined in Theorem 3.2 (specifically, these
are the ideals pZ[i] with p ≡ 3 mod 4). For m > 3 and p ∈ P we first claim we
have an injection of PSL(2; Z[i]/p j ) into Ω(m, 1; Z/p j Z). For this, we argue as
follows. First, there exists a quadratic form B4 defined over Q of signature (3, 1)
and an injection
PSL(2; Z[i]) −→ PSO0 (B4 ; Z).
For each prime p = pZ[i], this induces isomorphisms
PSL(2; Z[i]/p j ) −→ Ω(B4 ; Z/p j Z).
For j = 1, this can be found in [29]. For j > 1, this is established by an argument similar to that used in the proof of the equivalence of PSL(2; Z/p j Z) and
Ω(2, 1; Z/p j Z) in proving Theorem 1.1. Extending the form B4 from Q4 to Qm+1
for m > 3 by the identity produces injections
Ω(B4 ; Z/p j Z) −→ Ω(m, 1; Z/p j Z).
11
In particular, we can view
PSL(2; Z[i]/p j ) < Ω(m, 1; Z/p j Z)
for all m > 3, all j, and all p ∈ P as claimed. Since we have already shown in the
proof of Theorem 1.1 that π1 (M) surjects finite groups containing Ω(m, 1; Z/p j Z)
for all but finitely many primes, Theorem 3.2 (ii) with S = Z[i] and Theorem 2.3
(ii) now complete the proof.
4
Proof of Theorem 3.2
Throughout this section p will be an odd prime. For any ring R let M(2; R) be
the algebra of 2 × 2 matrices over R. The Lie algebra sl(2; F p ) of SL(2; F p ) consists of traceless matrices: sl(2; F p ) = {X ∈ M(2; F p ) | X11 = −X22 }. The adjoint action of SL(2; F p ) on sl(2; F p ) is the action by conjugation. As a vector
space sl(2; F p ) has a natural SL(2; F p )–invariant bilinear form, the Killing form
B defined by B(X,Y ) = Tr(XY ). The associated quadratic form QB (defined by
B(X, X) = 2QB (X)) is thus also invariant. Explicitly, for X,Y ∈ sl(2; F p ):
B(X,Y ) = 2X11Y11 + X12Y21 + X21Y12 ,
2
QB (X) = X11
+ X12 X21 .
Lemma 4.1. There is a short exact sequence
1 −→ sl(2; F p ) −→ SL(2; Z/p2 Z) −→ SL(2; F p ) −→ 1.
The conjugation action of SL(2; F p ) on sl(2; F p ) induced by this sequence is the
adjoint action.
Proof. The inclusion Z/pZ → Z/p2 Z is given by a 7→ pa. It induces an inclusion
M(2; Z/pZ) → M(2; Z/p2 Z) given by X 7→ pX.
Reduction modulo p induces the surjection π : SL(2; Z/p2 Z) → SL(2; F p )
whose kernel is clearly
ker(π) = {I + pX ∈ M(2; Z/p2 Z) | det(I + pX) = 1} .
Now det(I + pX) = 1 + p Tr(X) + p2 det(X) = 1 + p Tr(X) since we are in Z/p2 Z,
so we can rewrite:
ker(π) = {I + pX | X ∈ sl(2; F p )} .
The equation (I + pX)(I + pY ) = I + pX + pY now shows that the map X → I + pX
is an isomorphism of the additive group sl(2; F p ) to ker(π). The final sentence of
the lemma is clear.
12
Lemma 4.2. The number of SL(2; Z/p2 Z)–conjugacy classes in sl(2; F p ) (ie, orbits of the adjoint action of SL(2; F p )) is exactly (p + 2), as listed in the following table. In the table n represents a fixed quadratic non-residue in F p and “qr”
is short for quadratic residue (i.e., a square). Each of rows 2 and 3 represents
(p−1)/2 conjugacy classes, as Q = QB (X) runs respectively through the quadratic
residues and non-residues in F p − {0}.
description
size
# classes
trivial
1
1
anisotropic qr
p(p + 1)
(p − 1)/2
anisotropic non-qr
p(p − 1)
(p − 1)/2
isotropic qr
(p2 − 1)/2
1
isotropic non-qr
(p2 − 1)/2
1
representative
00
0 0
01
Q 0
01
Q 0
00
1 0
00
n0
Proof. We will prove this in several steps.
x
y
Step 1. Any z −x ∈ sl(2; F p ) is SL(2; F p )–equivalent to a matrix of the form
0 y0
0
z
0 .
To see this, note first that
a
c
b
d
x
z
y
−x
d
−c
−b
a
=
(1 + 2bc)x + bdz − acy
∗
∗
∗
,
so we want to solve the equations ad − bc = 1 and (1 + 2bc)x + bdz − acy = 0 for
a, b, c, d.
• If y 6= 0 choose b = 0, a = d = 1 and solve x − cy = 0 for c.
• If y = 0 and z 6= 0 choose a = 0, b = −c = 1 and solve for d.
• If y = z = 0 choose 2bc = −1, a = 1 and solve for d.
x
y
x
y
0
Step 2. If Q = Q z −x 6= 0 then z −x is SL(2; F p )–equivalent to Q
We have shown we can assume x = 0. Then
a b
0 y
d
−b
bdz − acy
= d 2 z − c2 y
c d
z 0
−c
a
a2 y − b2 z
acy − bdz
.
1
0
.
(3)
2
2
Since
Q =yz 6= 0we have y, z 6= 0 so a y − b z = 1 can be solved for a, b. Then
a b
a
b
c d = bz ay does what is required.
13
x
y
x
y
Step 3. Excluding the zero-element, if Q z −x = 0 then z −x is SL(2; F p )–
0 0
0 0
equivalent to exactly one of 1 0 or n 0 , where n is a fixed quadratic nonresidue.
We can assume x = 0. If z = 0 we conjugate by an element with a = 0 to get
y = 0. Thus we can assume x = y = 0 and z 6= 0. Now looking at equation (3), one
sees that if x = y = 0 then z can be changed only by squares.
Step 4. It remains to verify the sizes of the conjugacy classes.
Foreach class in row 2 or 3 we must simply count the number of elements
x
y
2
z −x with x + yz = Q. Here Q 6= 0. If Q is a quadratic non-residue then we
must have yz 6= 0, so for each of p choices of x and each of p − 1 choices of y 6= 0
we get a unique z. There are therefore p(p − 1) elements in the class. A similar
count gives p(p + 1) elements if Q is a residue.
If Q = 0 it is easier to work out the isotropy group of a representative of
the class. For
in our normal form x = y = 0 the isotropy group con −1an element
d
0
sists of all c d with d 2 = 1. This clearly has size 2p so the class has size
| SL(2; F p )|/2p = (p2 − 1)/2.
We now investigate the SL(2; F p )–classes of proper non-trivial subgroups in
sl(2; F p ). The group sl(2; F p ) itself has order p3 .
We first consider the subgroups of order p. Using Lemma 4.2 it is clear there
are three classes. Namely
I. Isotropic lines. Each isotropic line has (p − 1)/2 isotropic qr elements and
(p − 1)/2 isotropic non-qr elements.
There are p + 1 such lines in this class.
0 0
A representative is the line y 0 | y ∈ F p .
R. Anisotropic qr lines. Each such line has exactly two elements in each anisotropic
qr conjugacy class. There
are p(p + 1)/2 such lines in this class. A repre 0 y sentative is the line y 0 | y ∈ F p .
N. Anisotropic non-qr lines. Each such line has exactly two elements in each
anisotropic non-qr conjugacy class. There are p(p − 1)/2 such lines in this
0 y
class. A representative is the line ny 0 | y ∈ F p .
Next, we determine the conjugacy classes of subgroups of order p2 , i.e., planes.
Since the Killing form B is nonsingular, the orthogonal complement of such a plane
with respect to B will be a line, and vice versa, so we can classify planes up to
14
conjugacy by the conjugacy classes of their orthogonal complements. There are
therefore three classes of planes:
I ⊥ . Orthogonal complements of isotropic lines. A representative such plane is
x 0 I ⊥ = y −x | x, y ∈ F p .
The Killing form is degenerate on this plane, with nullspace I. This nullspace
contains all isotropic elements of the plane and the remaining elements consist of 2p elements from each anisotropic qr conjugacy class. The plane has
no anisotropic non-qr elements. There are p + 1 of these planes.
R⊥ . Orthogonal complements of anisotropic qr lines. A representative such plane
is
x y R⊥ = −y −x | x, y ∈ F p .
Such a plane has exactly 2p − 2 isotropic elements, which, together with 0,
form two isotropic lines (in R⊥ the lines x = y and x = −y). For any Q 6= 0
there are exactly p − 1 elements X ∈ R⊥ with QB (X) = Q. Thus such a plane
intersects every conjugacy class in sl(2; F p ). There are p(p + 1)/2 of these
planes.
N ⊥ . Orthogonal complements of anisotropic non-qr lines. A representative such
plane is
x
y
N ⊥ = −ny −x | x, y ∈ F p .
Such a plane has no isotropic elements and for any Q 6= 0 it has p + 1 elements with QB (X) = Q. There are p(p − 1)/2 of these planes.
We note for future reference
Lemma 4.3. Any plane of type R⊥ is elementwise conjugate in SL(2; Z/p2 Z) to
sl(2; F p )
Proof of Theorem 3.2 (i). We will show that the trivial subgroup is fixed point
equivalent to any anisotropic qr line R. It suffices to show that the only finite
dimensional representation of SL(2; Z/p2 Z) without an R–fixed vector is the trivial representation. Given such a representation, each subgroup H containing R will
also have no fixed vector. We will use this information for the subgroups H of type
R, I ⊥ , R⊥ , N ⊥ , and sl(2; F p ) to show the representation must be trivial.
To begin, the sum of the character χ of a representation over the non-zero elements of a line in sl(2; F p ) will only depend on the conjugacy class of the line, and
thus give numbers that we shall call XI (χ), XR (χ), XN (χ), depending on whether
15
the line is isotropic, anisotropic qr, or anisotropic non-qr. Also, let X0 (χ) be the dimension of the representation; this is χ evaluated on the trivial element in sl(2; F p ).
If H is a subgroup of sl(2; F p ), then the sum of χ over the elements of H gives |H|
times the dimension of the fixed space of the representation restricted to H, hence
zero under our assumption that H has no non-trivial fixed points. Since H is a
union of lines that are disjoint except at 0, this then gives an equation of the form
X0 (χ) + IH XI (χ) + RH XR (χ) + NH XN (χ) = 0 .
Here the coefficients IH , RH , NH are the number of lines of each type in H. By our
discussion above, these numbers for the subgroups of interest to us are:
IH
RH
NH
H =R
0
1
0
H = I⊥
1
p
0
H = R⊥
2
(p − 1)/2 (p − 1)/2
H = N⊥
0
(p + 1)/2 (p + 1)/2
H = sl(2; F p ) p + 1 (p2 + p)/2 (p2 − p)/2
These five different types of subgroups containing R yield five linear equations in
the four unknown quantities X0 (χ), XI (χ), XR (χ), XN (χ). Since already the coefficient matrix of the first four equations,


1 0
1
0
1 1

p
0


1 2 (p − 1)/2 (p − 1)/2 ,
1 0 (p + 1)/2 (p + 1)/2
has nonzero determinant (namely −p2 ), the equations have only the trivial solution.
This implies X0 (χ) = 0, proving the representation is trivial, as desired.
Remark. By computing the character table of SL(2; Z/p2 Z) one can show that
there is no other fixed point equivalence in SL(2; Z/p2 Z) between non-conjugate
subgroups of sl(2; F p ).
Proof of Theorem 3.2 (ii). Let p be a prime ideal of Ok such that Ok /p = Fq is
a proper extension of F p and p > 3; that such a prime exists follows from the
Cebotarev Density Theorem. Consider the following inclusion of short exact sequences:
1
/ Vp
/ PSL(2; Z/p2 Z)
/ PSL(2; F p )
/1
1
/ Vp
/ PSL(2; O /p2 )
k
/ PSL(2; Fq )
/1
16
By Lemma 4.1 we already know the kernel Vp in the first sequence is sl(2; F p ) (the
transition from SL to PSL just factors by {±I} and does not affect the kernel).
Although we do not need it, we note that Vp = sl(2; Fq ). If p is principal,
p = (π), say, then we could argue as in the proof of Lemma 4.1. In general we can
replace k by its localization at p without changing the second exact sequence and
then p becomes principal, so the argument applies.
We claim that in PSL(2; Ok /p2 ) any element of Vp can be conjugated out of
Vp . We only need show this for the representatives of conjugacy classes given
in Lemma 4.2 and the claim is then a simple calculation using equation (3) with
b ∈ Fq − F p , a = d = 1, and c = 0.
The proof is now complete, since Lemma 4.3 gives elementwise conjugate
subgroups in Vp and we have just shown they are primitive in PSL(2; Ok /p2 ).
5
Locally symmetric manifolds and other generalities
5.1
R–rank 1 geometries
We shall denote by HYn the n–dimensional hyperbolic spaces modelled on Y ∈
{C, H, O} (where n = 2 when Y = O). The methods used to produce eigenvalue,
length, and primitive length equivalent manifolds extend with little fuss to complex,
quaternionic, and Cayley hyperbolic manifolds. We give the version for primitive
length.
Theorem 5.1. Let Γ be a torsion-free lattice in Isom(HYn ). Then there exist infinitely many pairs of finite covers of M = HYn /Γ, {M j , N j } such that
(a) L p (M j ) = L p (N j ),
(b) vol(M j )/ vol(N j ) → ∞.
Moreover, (a) and (b) hold for any finite volume Riemannian metric on M.
Proof. The argument we give breaks into a few cases. First, in most cases we have
the inclusion
PO0 (B4 ; Z) < GY,n (Z)
where B4 is the form from the proof of Theorem 1.2, GY,n is Q–algebraic, and
GY,n (R) with the analytic topology is Lie isomorphic to Isom(HYn ). For Y = C,
this fails only when n = 1, 2. For Y = H, when n ≥ 3, this is clear. The remaining cases of n = 1, 2 follows from the exceptional isometry between H1H and H4
together with the isometric inclusion of H1H into H2H . Finally, for Y = O, this follows from the isometric inclusion of H2H into H2O . For all these cases, as in the
17
proof of Theorem 1.2, an application of the Strong Approximation Theorem (cf
[30], [16]) in combination with the Cebotarev Density Theorem provides infinitely
many primes p such that Γ surjects onto certain finite groups G(S/p2 ) of Lie type
which contain PSL(2; Z[i]/p2 Z[i]). The proof is completed just as it was in the
proof of Theorem 1.2.
It remains to deal with Y = C and n = 1, 2. The case of n=1 is simply the
case of hyperbolic surfaces. Case 2 cannot be handled indirectly, and we must use
primitive pairs in the finite groups PU(2, 1; Ok /p2 ), where k/Q is an imaginary
quadratic extension of Q and p is a prime ideal of Ok . Selecting p such that Ok /p
is a quadratic extension of F p , we have the short exact sequence
1 −→ su(2, 1; Ok /p) −→ SU(2, 1; Ok /p2 ) −→ SU(2, 1; Ok /p) −→ 1,
where su(2, 1; Ok /p) is the Lie algebra of SU(2, 1) over the field Ok /p. With the
inclusions
Ω(2, 1; F p ) < SU(2, 1; Ok /p),
sl(2; F p ) < su(2, 1; Ok /p),
The subgroups sl(2; F p ) and R⊥ are elementwise conjugate in SU(2, 1; Ok /p2 )
where R⊥ is a 2–plane from Lemma 4.3. It is straightforward to verify that the
pair satisfies the additional requirements needed for the primitive case.
Our methods also produce eigenvalue equivalent covers for all of these groups
as well. In addition, for sufficiently large n, we can produce covers which are
both primitive length and eigenvalue equivalent; here n ≥ 5 and Y can be R, C, or
H. To do this, by [13] Proposition 4 Window 2 for n ≥ 5 we can arrange for the
simple groups of orthogonal type to contain a copy of (P) SL(3; F p ) which contains
a group of the type given in Theorem 2.7.
5.2
Locally symmetric manifolds
Length and eigenvalue equivalent covers As is clear from this discussion (and
the generality of the Strong Approximation Theorem in [30] and [16]) our methods
also apply to lattices in every non-compact higher rank simple Lie group. The
discussion given at the end of §5.1 also applies in this setting to arrange for the
finite groups of Lie type occurring in Strong Approximation to contain a group a
copy of (P) SL(3; F p ).
Primitive length equivalent covers Construction of primitive length equivalent
covers over a fixed locally symmetric manifold is more subtle since in many settings the associated fundamental group fails to have the needed condition on maximal cyclic subgroups. It seems interesting to try to weaken the condition on maximal cyclic subgroups to produce examples in this setting.
18
6
6.1
Final Remarks
Relations among length, primitive length, and eigenvalue equivalence
Example A. Let M be a closed surface of genus at least 2 equipped with a hyperbolic metric. Let G be the alternating group Alt(4) and H and K the elementwise conjugate pair described in §2.3. Then, given a surjection π1 (M) → G, let
γ ∈ π1 (M) map to a ∈ G and correspond to a primitive closed geodesic in M (there
are infinitely many primitive elements mapping to any element of G). The nonprimitive geodesic of M corresponding to γ has four lifts to MH , two primitive and
two not, and it has three lifts to MK , all non-primitive. Of course, there might
accidentally be some unrelated primitive geodesic in MK of the right length, but
for a generic hyperbolic metric, and a homomorphism to G that factors through a
free group this does not happen and MH and MK are not primitive length equivalent. Indeed, assume γ is the shortest closed geodesic on M and every other closed
geodesic has much larger length. Then if γ maps to a, one can see that MH and MK
are not primitive length equivalent.
Example B. Eigenvalue equivalent surfaces obtained from Theorem 3.1 using the
trivial subspace {0} and any proper subspace of Fnp generically produce examples
which are not length equivalent. In particular, eigenvalue equivalence need not
imply length or primitive length equivalence.
It seems plausible that length equivalent hyperbolic examples constructed from
Theorem 3.1 using Fnp and any nontrivial subspace of Fnp will generically fail to be
eigenvalue equivalent, but this is more subtle. Using the results of Zelditch [31]
it is easy to see that for a hyperbolic manifold M m of sufficiently high dimension
this approach will give length equivalent but not eigenvalue equivalent examples
for generic (not necessarily hyperbolic) deformations of the metric on M. Using
G = Alt(4) this allows one to find such examples in dimensions m ≥ 3.
All of our examples of primitive length equivalence are also examples of eigenvalue equivalence.
Question 6.1. Are two primitive length equivalent hyperbolic manifolds necessarily eigenvalue equivalent?
6.2
Complex lengths
All our results for equal length sets actually produce manifolds which have the
same complex length sets. Recall that the complex length of a closed geodesic γ
in a Riemannian m–manifold is a pair (`(γ),V ) where `(g) is the length of γ and
19
V ∈ O(m − 1) is determined by the holonomy of γ. The complex length spectrum is
the collection of such complex lengths with multiplicities, and the complex length
set forgets multiplicities as before. The point is that Theorem 2.3 gives manifolds
with the same complex length sets, just as Sunada’s theorem gives equal complex
length spectra. See [23] for more on the complex length spectrum.
6.3
Commensurability
The known methods of producing isospectral or iso-length spectral hyperbolic
manifolds result in commensurable manifolds and it is an open question as to
whether this is always the case. By construction, the eigenvalue and (primitive)
length equivalent hyperbolic manifolds constructed here are also commensurable.
Question 6.2. Let M1 and M2 be eigenvalue (resp. length or primitive length)
equivalent closed hyperbolic manifolds. Are they commensurable?
There has been some recent activity on this question. It is shown that Question
6.2 has an affirmative answer in the length equivalent setting if the manifolds M1
and M2 are arithmetic hyperbolic 3–manifolds ([3]), or if the manifolds are even
dimensional arithmetic hyperbolic manifolds ([21]). Indeed, the results of [21]
apply to more general locally symmetric spaces. In contrast, [21] also exhibts
arbitrarily large collections of incommensurable hyperbolic 5–manifolds which are
length commensurable. The commensurability classes of these manifolds seem to
be the best candidates for producing a negative answer Question 6.2.
6.4
Infinite sets of examples
Our constructions show that there can be no uniform bound on the number of pairwise eigenvalue (resp. length or primitive length) equivalent, non-isometric manifolds. Thus a natural question is.
Question 6.3. Are there infinite sets of pairwise eigenvalue (resp. length or primitive length) equivalent, closed hyperbolic m–manifolds?
In the context of length equivalence a positive answer would follow if one
can find infinitely many mutually elementwise conjugate subgroups of finite index
in a finitely generated free group. C. Praeger pointed out to us that a slightly
stronger version of this question is listed as an open problem (Problem 11.71) in the
Kourovka Notebook [15]. It was asked there in the parallel context of Kronecker
equivalence of number fields. It seems likely that the answer to this question is
“no”, but the limited partial answers that are known involved considerable effort,
see [20].
20
6.5
Can one hear the size of a drum?
Mark Kac’s famous paper “Can one hear the shape of a drum” [9] is quoted in
many papers on isospectrality. Of course, the “drums” of his title were not closed
hyperbolic manifolds, but rather flat plane domains. The first pair of different
“drums” with the same sound (i.e., non-isometric isospectral plane domains) were
found in the 1990’s by Gordon, Webb, and Wolpert [5].
However, one might question whether the sounds of their drums D1 and D2 are
really indistinguishable. They comment: “... to produce the same sound (i.e., the
same frequencies with the same amplitudes) as would result from striking D1 at a
given point with a given (unit) intensity ... one must strike D2 simultaneously at
seven points with appropriate intensities”. A more obvious example of this issue
is a pretty example of S. Chapman [2]. Chapman reinterprets earlier discussion of
the Gordon–Webb–Wolpert examples in terms of paper folding and cutting, as is
familiar from making paper dolls. Of course, by cutting too much one can create
disconnected objects, and by this means Chapman derives from the Gordon–Webb–
Wolpert example the following simple example: D1 is the disjoint union of a unit
square and an isosceles right triangle with legs of length 2, and D2 is the disjoint
√
union of a 1 × 2 rectangle and an isosceles right triangle with legs of length 2.
This pair of domains is isospectral, but one can ask to what extent they really sound
the same.
A more honest example of equal sound might be the following: purchase three
identical drums and let D1 consists of one of them and D2 consist of the disjoint
union of the other two. It would be hard to distinguish D1 from D2 on hearing a
drummer strike either one once. This example suggests that eigenvalue equivalence
may have as much right as isospectrality to be interpreted as “same sound.”
In his paper Kac gave a proof that drums that sound the same have equal area,
but this was based on isospectrality. Revisiting this in the context of eigenvalue
equivalence we ask:
Question 6.4. Do there exist connected eigenvalue equivalent plane domains of
unequal area?
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21
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Department of Mathematics
University of Illinois at Urbana-Champaign
23
Urbana, IL 61801
email: clein@math.uiuc.edu
Department of Mathematics
California Institute of Technology
Pasadena, CA 91125
email: dmcreyn@caltech.edu
Department of Mathematics
Barnard College, Columbia University
New York, NY 10027
email: neumann@math.columbia.edu
Department of Mathematics
University of Texas
Austin, TX 78712
email: areid@math.utexas.edu
24